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Kilograms and Slugs

  1. Apr 19, 2007 #1
    Ok heres an easy one for you.

    we start with F=ma

    SI --> newtons=kilograms*gravity

    English --> pounds=slugs*gravity

    So now why if you weigh yourself on an english scale it is in pounds, but if you stand on a metric scale it is in kilograms? Ive solved plenty of F=ma problems throughout my college courses, but this question is something that just dosent seem clear to me. It seems that the units are wrong, shouldnt they be reading newtons & pounds or kilograms & slugs? I never even heard of a newton or a slug until I started studying engineering.
     
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  3. Apr 20, 2007 #2

    Gib Z

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    Well I have never dealt with pounds/slugs before but I don't see what the problem is. When we step on a scale that is metric, we get an idea of our mass by the kilogram value. However, if someone was to tell us how many Newtons we weighed, it would still give me an idea of my mass.

    So perhaps In England, thats exactly what their doing.
     
  4. Apr 20, 2007 #3

    Integral

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    It is all in how the lines are drawn on the scale. Americans weigh themselfs in lbs, Euorpeans , Mass themselfs in kg. What is the problem?

    BTW I think slugs is an archic unit, the offical unit is now lbm, pounds mass.
     
  5. Apr 20, 2007 #4
    well if a pound=slug*gravity then why do Euorpeans ignore gravity, and even offer such conversions such as 1 kilogram = 2.2046 pounds, they are different things being measured, weight vs mass
     
    Last edited: Apr 20, 2007
  6. Apr 20, 2007 #5

    rbj

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    okay, i guess that means non-relativistic speeds.

    not quite. Nt = kg * (1 m/s2)

    gravity on Earth is about 9.8 m/s2

    again, not quite. lbs = slug*(1 ft/s2)

    gravity on Earth is about 32.174 ft/s2

    a pound force is the force applied to accelerate a pound mass at the rate of 32.174 ft/s2 which happens to also be the acceleration of gravity. so if a pound mass is resting on the Earth, it pushes down with a force of one pound force.

    a slug would weigh 32.174 lbs. that means it's 32.174 pound masses.
     
    Last edited: Apr 20, 2007
  7. Apr 20, 2007 #6

    Gib Z

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    Abit Pedantic, but F=ma is valid for all constant accelerations if m is taken to be relativistic mass rather than rest mass.
     
  8. Apr 20, 2007 #7

    FredGarvin

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    The whole thing with Lbm is that, for most engineering calculations, one doesn't have to worry about the variation in the gravitational field. This permits the direct interchange between a Lbm and an Lbf. In realtion to the SI units and definition of a Newton, the true definition in the elemental units, [tex]1 Lbf = 1 sl*g[/tex]

    If you want to use Lbm in F=ma, then you need, to be dimensionally correct, to introduce [tex]g_c[/tex] where

    [tex]g_c = 32.2 \frac{Lb_m*ft}{Lb_f*s^2}[/tex]

    and F=ma needs to be adjusted by

    [tex]F = \frac{m*a}{g_c}[/tex]

    You can see that when you assume that one is at the surface of the Earth, at the equator, blah blah blah, [tex]g[/tex] and [tex]g_c[/tex] cancel out directly and an 1 Lbf = 1 Lbm. Of course, if you ever take a class that deals with space regimes like a rocket propulsion class I took, if you want to work in english units, you can't lose [tex]g_c[/tex]

    Personally, I hate dealing with it but I work for a company that has been around for a long time and a lot of the work done in the past directly interchanges the two units. Most people don't even take the time to distinguish between Lbf and Lbm.
     
  9. Apr 20, 2007 #8

    D H

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    Newton's second law does not say force is the product of mass and acceleration. Rather, it says force is proportional to this product: F=kma. The metric system was designed to make that constant of proportionality exactly one, or more precisely, 1 Ns2/kg/m:

    [tex]F = 1\frac{{\text Ns}^2}{\text{kg}\cdot\text{m}} m\,a[/tex]

    We drop the 1 and the units conversion for convenience to write F=ma. We can do the same in English units so long as mass is expressed in slugs, acceleration in ft/sec2, and force in pounds force. The constant of proportionality cannot be ignored when mass is in pounds mass.

    Slugs are still in use and are an official derived unit. Moments of inertia are almost always expressed in English units as slugs ft2, for example.

    We use "weight" informally to mean "mass". When we say something "weighs" 6 pounds, we really mean it has a mass of 6 pounds mass. That 6 pound mass will still have a mass of 6 Lbm on the moon but will only weigh about 1 lbf.

    A bit off-topic, the concept of relativistic mass was dropped in part because using F=ma is not valid. The relativistic mass changes depending on the direction of the force versus the direction of travel.
     
  10. Apr 20, 2007 #9

    russ_watters

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    This is really the point. It doesn't necessarily have anything to do with the actual physics definition of the words, it is just what has become the popular convention. Technically speaking, you can't get mass from a spring-scale directly.
     
  11. Apr 20, 2007 #10
    got it. seems like it's just a common language vs. scientifically correct terms issue then.
     
  12. Apr 20, 2007 #11

    rbj

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    speaking of pendantic. it doesn't work that way, Gib. F=ma is just not correct, even if m is relativistic mass, for speeds close to c.

    but F = dp/dt is correct, high speeds or slow.
     
  13. Apr 20, 2007 #12

    Gib Z

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    Yes but doesn't dp/dt yield ma when acceleration is constant? I know having a constant acceleration has numerous troubles that have to be taken into account near c, but still.
     
  14. Apr 20, 2007 #13
    What's really messed up is that there are digital scales that measure in kilograms--some of them measure weight and convert it to mass! How pompous they are to assume that we're weighing ourselves in a constant, 9.8 m/s^2 gravitational field!
     
  15. Apr 20, 2007 #14

    rbj

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    no, but dp/dt becomes ma when the mass is constant. kicking into LaTeX:

    [tex] F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m \frac{dv}{dt} + v\frac{dm}{dt} [/tex]

    the first term of the sum you recognize and the second goes to zero in the context where mass is constant, which isn't the case when accelerating something close to c.
     
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